Question 174002
your equation is:
1/2[log(base4)(x+1) + 2log(base4)(x-1)] +6log(base4)(x)
this looks to me like:
{{{(1/2)*((log(4,x+1)) + (2*log(4,x-1))) + 6*log(4,x)}}} 
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you have a couple of concepts to work with:
these are:
{{{log(4,a) + log(4,b) = log(4,a*b)}}}
{{{c*log(4,a) = log(4,a^c)}}}
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we will remove the inner parentheses first and work outwards.
you have:
{{{2*log(4,x-1)}}}
this becomes:
{{{log(4,(x-1)^2)}}}
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working our way outwards, you have:
{{{log(4,x+1) + log(4,(x-1)^2)}}}
this becomes:
{{{log(4,(x+1)*(x-1)^2)}}}
working our way outwards again, you have:
{{{(1/2)*(log(4,(x+1)*(x-1)^2))}}}
this becomes:
{{{log(4,((x+1)*(x-1)^2)^(1/2))}}}
we now have in total:
{{{log(4,((x+1)*(x-1)^2)^(1/2)) + 6*log(4,x)}}}
{{{6*log(4,x) = log(4,x^2)}}}
so we have:
{{{log(4,((x+1)*(x-1)^2)^(1/2)) + log(4,x^6)}}}
and this becomes:
{{{log(4,(((x+1)*(x-1)^2)^(1/2)*x^6))}}}
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i checked the answer and it appears to be good.
i did this by solving both equations for a given value of x.
i used x = 2.
i got the same answer both ways (the original equation and the condensed equation).
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you can solve this as well if you wish to prove the answer is good.
you can use your calculator by converting to base 10 as follows:
{{{log4,x) = log(10,x)/log(10,4)}}}
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